Types of Beams, design example and design sheet in excel file download

1. Rectangular Beams (Rec)

• Shape: A simple rectangular cross-section.
• Applications: Frequently used in buildings, bridges, and other structures due to ease of design and construction.
• Advantages: Easy to fabricate and calculate; standard dimensions readily available.
• Disadvantages: May require larger dimensions for higher loads, leading to heavier designs.

2. Circular Beams (Circular or Cylindrical)

• Shape: A circular or cylindrical cross-section.
• Applications: Used in structures requiring symmetrical aesthetics, such as columns doubling as beams or specific industrial applications.
• Advantages: Equal resistance in all directions (isotropic in bending), aesthetically pleasing.
• Disadvantages: Less efficient in terms of material usage compared to rectangular beams.

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Design Example: Rectangular Beam

Problem: Design a simply supported rectangular beam of span L = 6 m, subjected to a uniformly distributed load of w = 10 kN/m. The allowable bending stress is f_b = 10 MPa.

Steps:

1. Calculate Maximum Bending Moment (M_max): M_max = (wL²) / 8
   Substituting the values:
   M_max = (10 × 6²) / 8 = (10 × 36) / 8 = 45 kN·m = 45,000 N·m = 45,000,000 N·mm.
2. Section Modulus (Z) Requirement: Z = M_max / f_b
   Substituting the values:
   Z = 45,000,000 / 10,000,000 = 4.5 m³ = 4,500,000 mm³.
3. Select Dimensions of Beam: For a rectangular section, Z = (b d²) / 6.
   Assume b = 250 mm (breadth).
   Solve for d (depth): 4,500,000 = (250 × d²) / 6
   d² = (4,500,000 × 6) / 250 = 27,000,000 / 250 = 108,000
   d = √108,000 ≈ 328.63 mm.
   Round d to a practical value:
   d = 350 mm.
4. Verify the Design: Recalculate Z = (b d²) / 6:
   Z = (250 × 350²) / 6 = (250 × 122,500) / 6 = 5,102,083.33 mm³.
   Check if Z > required Z:
   5,102,083.33 mm³ > 4,500,000 mm³ (Safe!)

Final Beam Dimensions:
Breadth b = 250 mm, Depth d = 350 mm.

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Design Example: Circular Beam

Problem: Design a simply supported circular beam of span L = 6 m, subjected to the same uniformly distributed load of w = 10 kN/m and the allowable bending stress f_b = 10 MPa.

Steps:

1. Calculate Maximum Bending Moment (M_max): M_max = (wL²) / 8
   Substituting the values:
   M_max = (10 × 6²) / 8 = 45 kN·m = 45,000,000 N·mm.
2. Section Modulus (Z) Requirement: Z = M_max / f_b
   Substituting the values:
   Z = 45,000,000 / 10,000,000 = 4.5 m³ = 4,500,000 mm³.
3. For Circular Section: Section modulus Z = (π d³) / 32.
   Solve for d (diameter): 4,500,000 = (π × d³) / 32
   Rearranging:
   d³ = (4,500,000 × 32) / π = 144,000,000 / 3.1416 = 45,873,572.8
   d = ∛45,873,572.8 ≈ 357.7 mm.
   Round d to a practical value:
   d = 375 mm.
4. Verify the Design: Recalculate Z = (π d³) / 32:
   Z = (π × 375³) / 32 = (3.1416 × 52,734,375) / 32 = 5,203,703.7 mm³.
   Check if Z > required Z:
   5,203,703.7 mm³ > 4,500,000 mm³ (Safe!)

Final Beam Diameter:
Diameter d = 375 mm.

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